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Question Number 217732 by ArshadS last updated on 19/Mar/25

$$\{\begin{array}{l}x+y=xy\\ x^2+y^2=25\end{array};x^4+y^4=?$$

Commented by Ghisom last updated on 19/Mar/25

$$x^4+y^4=571\pm 4\sqrt{26}$$

Answered by Wuji last updated on 19/Mar/25

$$S=x+y,P=xy$$$${\left(xy\right)}^2=P^2=S^2$$$$x^4+y^4={(x^2+y^2)}^2-2{\left(xy\right)}^2$$$$x^4+y^4={\left(25\right)}^2-2S^2=625-2S^2$$$$x^2+y^2=S^2-2P\Rightarrow S^2-2S=25$$$$S^2=2S+25$$$$\Rightarrow x^4+y^4=625-2(2S+25)=625-4S-50$$$$x^4+y^4=575-4S$$$$from{S}^2-2S-25=0$$$$S=\frac{2\pm \sqrt{{(-2)}^2-4\times (-25)}}{2}=\frac{2\pm \sqrt{4+100}}{2}=\frac{2\pm \sqrt{104}}{2}$$$$S=\frac{2\pm 2\sqrt{26}}{2}=1\pm \sqrt{26}$$$$S=1+\sqrt{26}S>0$$$$\Rightarrow x^4+y^4=575-4(1+\sqrt{26})$$$$x^4+y^4=575-4-4\sqrt{26})=571-4\sqrt{26}$$$$\therefore x^4+y^4=571-4\sqrt{26}$$

Commented by ArshadS last updated on 22/Mar/25

$$Thanksalot!$$

Answered by Rasheed.Sindhi last updated on 19/Mar/25

$$x^2+y^2=25$$$$\Rightarrow {(x+y)}^2-2xy=25$$$$\Rightarrow {(x+y)}^2-2(x+y)=25$$$$\Rightarrow {(x+y)}^2-2(x+y)-25=0$$$$x+y=xy=\frac{2\pm \sqrt{4+100}}{2}=1\pm \sqrt{26}$$$${(x^2+y^2)}^2={25}^2$$$$x^4+y^4+2{\left(xy\right)}^2=625$$$$x^4+y^4=625-2{(1\pm \sqrt{26})}^2$$$$=625-2(1+26\pm 2\sqrt{26})$$$$=625-54\mp 4\sqrt{26}$$$$=571\mp 4\sqrt{26}$$$$=571\pm 4\sqrt{26}$$

Commented by ArshadS last updated on 22/Mar/25

$$Thankssir!$$

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